Voltage potential in PEM electrolyzers

[HelloCthulhu]

I've been studying PEM fuel cells for a while and wondered if they could be used for electrolysis. I was surprised at first to find out the technique had been used for decades, but after watching this animation the dynamics make much more sense:

https://en.wikipedia.org/wiki/Polym...lectrolysis#/media/File:PEM_Elektrolyse_5.gif

However, my understanding of electronegativity, voltage, and efficiency is still lacking for both fuel cells and electrolyzers. Wiki ranges the cell voltage for an electrolyzer at 1.75-2.20 V and the range of efficiency at 57-69%. I know that splitting a mole of liquid water to produce a mole of hydrogen at 25°C requires 285.8 kJ of input energy for a battery—237.2 kJ as electricity and 48.6 kJ as heat. And that cell voltage is 1.48V at standard atm and temperature:

$$E_o=\frac{(∆_f H)}{zF}=\frac{285,840 J/mol}{2*96,485 C/mol}=1.48Volts/cell$$

The question I have is regarding half-cell potential and electronegativity. In terms of voltage, why is the hydrogen attracted to the cathode side of the cell? How does this affect the efficiency of the cell? I'm trying to use a galvanic cell as reference, but, since hydrogen isn't losing it's proton in solution the same way copper and zinc do in the example, I'm still finding myself confused.

https://en.wikipedia.org/wiki/Galvanic_cell

 

[Lord Jestocost]

To clarify:

The minimum necessary voltage to start water electrolysis under standard conditions is ≈ 1.23 V.
The voltage necessary to start water electrolysis without withdrawing heat from the surroundings under standard conditions is ≈ 1.48 V.

 

[HelloCthulhu]

Thanks for your input! But I'm still confused about how voltage on the anode side is related to the increased electronegativity on the cathode side of the cell. Any thoughts on what forces the hydrogen protons to separate from oxygen in terms of voltage? It might be something simple, but I'm just not seeing the connection yet.

 

[HelloCthulhu]

I think the relationship that confuses me is between the dissociation energy necessary to split 1 mole of water and the electronic potential between the electrodes. Hopefully researching electrode kinetics will put me on the right track!

 

[Lord Jestocost]

The decomposition of a water molecule H₂O – to use the simplest picture – means to transfer two electrons from the oxygen ion O^-- back to two hydrogen ions H⁺. To accomplish this you need a certain amount of energy input.

Neglecting every detail, equilibrium thermodynamics says: Such ions are always present, whether in liquid water or at metal surfaces in contact with water. The respective concentrations depend on the given conditions.

You can now use “electrolysis cells” to separate these ions under the influence of an applied electric field (in case of PEM cells you force H⁺ ions which exist at the three phase boundary “anode metal/water (liquid or gaseous)/PEM membrane” to migrate through the membrane to the cathode region, O^-- ions "remain" at the three phase boundary in the anode region).

At the end, if the applied voltage is high enough, you can even “extract” the electrons from the oxygen ions at the anode region and “transfer” them to the hydrogen ions which arrive at the cathode. By this, you establish a continuous decomposition of water molecules.

Have a look at http://www1.lsbu.ac.uk/water/electrolysis.html for details with respect to classical water electrolysis.